I was trying to put together all the things I've read in the last couple of days and I realized that based on my current knowledge of the standard model and the way nuclear fission works, I'm not able to understand why would any number of nuclei packed together by the strong nuclear force, no matter how tight, release energy when split.

This took me on an interesting rabbit hole and then I found this answer. Leaving aside the rant, it made me wonder, do we really understand why the splitting releases energy, or we just know it does based on observations.

The way I see things now, since the nuclear forces are so strong, their tendency should be to always bring nuclei together, even if bombarded with an extra neutron. The extra neutron should stick, or if the kinetic force is strong enough, then break the binding and split the nucleus, but certainly not releasing energy, losing mass and shooting 1-3 neutrons in the process.

So, why is my expectation wrong, since experimentally it obviously is.

  • $\begingroup$ Plenty of nuclei fall apart for other reasons too. If being apart is a lower energy configuration than it can happen. $\endgroup$
    – Jon Custer
    Jan 4, 2019 at 0:55
  • $\begingroup$ I vaguely recall that the inverse of the fine structure constant comes into play here. The element 137 is sometimes called Feynman-ium because he put forward the idea that somewhere in the vicinity of 1/fine_structure protons the nucleus could not be stable ... $\endgroup$
    – Paul Young
    Jan 4, 2019 at 0:58
  • $\begingroup$ "The way I see things now, since the nuclear forces are so strong, their tendency should be to always bring nuclei together" - but the strong nuclear force is short-ranged while the electrostatic repulsion between the protons is not. Is this what you're overlooking? $\endgroup$
    – Hal Hollis
    Jan 4, 2019 at 1:09

1 Answer 1


You are essentially asking why the binding energy of nuclei decreases beyond a certain point in the periodic table (the binding energies being strongest around iron). According to the liquid drop model (which is pretty accurate), there are two effects that play into this. The simpler effect is simply that all the protons in the nucleus repel one another, and this electromagnetic repulsion is long range, while the strong attraction is only short range.

Because it has such limited range, the residual strong interaction produces a binding energy proportional to the number of nearest-neighbor nucleon pairs. For large nuclei, this results is an essentially constant binding energy per nucleon. Almost every nucleon is fully surrounded by the neighbors, and so the total binding energy is proportional to the number of nucleons $A$. For smaller nuclei, there are a substantial number of nuclei on the surface, which is why smaller nuclei are less bound.

However, the electrostatic energy of all the protons in the nucleus does not increase linearly with with number of constituents. Instead, it increases as $Z^{2}$, where $Z$, the atomic number is the number of protons. So at some point, adding more protons to the nucleus is going to actually decrease the binding energy per nucleon, and this is the point at which fission becomes an exothermic reaction.

This does not answer the question of why we cannot simply add more and more neutrons to the nucleus, however. The reason for that is quantum mechanical. The neutrons (like the protons) are fermions and thus obey the exclusion principle. If we have more neutrons than protons, the Fermi/zero-point motion of the neutrons will be greater, while their binding potentials will be the same. If the nucleus becomes too neutron rich, it then becomes energetically favorable for a neutron to decay into a proton.

For small $Z$, when the electrostatic repulsion effect is small [because the fine structure constant $\alpha\approx1/137$, while the strong coupling is ${\cal O}(1)$], the most stable configurations have basically equal numbers of protons and neutrons, because this minimizes the energy of the nucleons' Fermi motion. As the electrostatic effect becomes more significant, the line of stability shifts toward having more neutrons than protons, but eventually (in the neighborhood of $z\sim1/\alpha$), you cannot add more of either kind of nucleon with decreasing the stability.

  • $\begingroup$ Yes. The pairing energy of adding a thermal neutron to a nucleus with an odd number of neutrons makes it oscillate like a liquid drop, protons get far apart and split, and the fragments then repel each other with the Coulomb force. This gives them a large kinetic energy which becomes heat when slowed down in the solid. $\endgroup$
    – user137289
    Jan 4, 2019 at 8:27
  • $\begingroup$ Thanks for this very detailed answer, it solves my doubt and puts me on path for further study. $\endgroup$
    – Rad'Val
    Jan 4, 2019 at 9:16

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